# Direction of the fastest rate of change of a function (Solution inclu)

1. Apr 18, 2013

### mrcleanhands

1. The problem statement, all variables and given/known data
Find all the points at which the direction of the fastest rate of change of the function $f(x,y)=x^{2}+y^{2}-2x-4y$ is i+j

2. Relevant equations

3. The attempt at a solution
The direction of the fastest change is $\nabla f(x,y)=(2x-2)i+(2y-4)j$, so we need to find all points (x,y) where $\nabla f(x,y)$ is parallel to i+j.

$\Longleftrightarrow(2x-2)i+(2y-4)j=k(i+j)$ $\Longleftrightarrow$ $k=2x-2$ and $k=2y-4$

Ok here's where I start to get lost. How can k = two different things at the same time?

The solution continues:
Then $2x-2=2y-4$ $\Longrightarrow y=x+1$, so the direction of the fastest change is i+j at all points on the line y=x+1

2. Apr 18, 2013

### LCKurtz

Right. You can't have $k$ equal two different things. What you do have to do is figure out what relation $y$ and $x$ must satisfy so that you have only one value $k$ needed. That requires that $2x-2=2y-4$.

3. Apr 18, 2013

### HallsofIvy

Staff Emeritus
The same way 4 can be equal to 2+ 2 and 3+ 1 "at the same time"= they are different ways of writing the same thing. That is why you can, as you say, write 2x- 2= 2y- 4.

4. Apr 18, 2013

### mrcleanhands

Ok, it just clicked.